A flywheel is a solid, rotating disk that is sometimes used to
store energy. The moment of inertia for a solid disk is 1/2MR^2
where M is the mass and R is the radius. If a
flywheel weighs 15kg and has a radius of 0.8m, determine the
following.
A) If a point on the outside of the disk is moving at 6 m/s, what
is the angular velocity of the flywheel?
B) If the flywheel starts from rest and accelerates at a constant
rate for 60 s to bring it up to the angular velocity found in Part
A, what is the angular acceleration?
C) If the angular acceleration in Part B is generated by a constant
force applied tangentially to the outside edge of the disk (as
shown below), what is the magnitude of the force?
Part A)
w = v/r
w = (6)/(.8)
w = 7.5 rad/s
Part B)
Apply wf = wo + (alpha)t
7.5 = 0 + alpha(60)
alpha = .125 rad/s^2
Part C)
Apply Torque = I(alpha) and FL, Thus
(.5mr^2)(.125) = F(r)
.5(15)(.8)(.125) = F
F = .75 N
a)
angular velocity, w = v / r
= 6 / 0.8
= 7.5 rad./ s
b)
Angular acceleration, alpha = (7.5 - 0) / 60
= 0.125 rad./s^2.
c)
F = torque / r
= I x alpha / r (where, I = moment of Inertia of disk.)
= (1/2) x m x r^2 x 0.125 / r
= 0.5 x 15 x 0.8 x 0.125
= 0.75 N.
A)omega= V/R=6/0.8=7.5 rad/s
B)angular cceleration=7.5/60=0.125 rad/s^2
C)T=I*Alfa
I=0.5*M*R^2=0.5*15*(0.8)^2=4.8 kg m^2
T=0.125*4.8=0.6 N-m
T=F*R
F=T/R=0.6/0.8=0.75 N ans
A) v= Rw, w = 6/.8 =7.5 rad./s
B) because w = w0+at so a = w/t = .125rad/s^2
C)because F*R = I*a so F = I*a/R =.75n
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